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Alisiya [41]
3 years ago
13

4. Julie purchased tickets to a Mets baseball game from the Mets website. Each ticket costs $19

Mathematics
1 answer:
Shkiper50 [21]3 years ago
6 0
Yhhehehehehehehehedhdhdhdh
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Yo can anybody answer this for me
Zanzabum

Answer: m=9

Step-by-step explanation: add 16 to both sides, Which gives you 36.Now you have 4m=36 ,so you divide 4 on both sides which gives you 9 .So it is m=4

8 0
3 years ago
Read 2 more answers
What will be the new reflected point across the x - axis of (-3, -3)?
Andreas93 [3]
The new reflected point across the x-axis of(-3,-3) will become (-3,3)
And the new reflected point across the y-axis of (-4,-4) will become (4,-4)
3 0
2 years ago
In this circle, the area of sector COD is 50.24 square units.
Bezzdna [24]

Answer:

Radius = 8 units

Length of arc AB = 8.37758 units

Step-by-step explanation:

The sector COD is 1/4 the size of the circle.

the vertex angle is 90° and a circle has 360° at the center.

So the area of sector COD/area of Circle = 90/360 = 1/4

This means the area of the circle is 4 * 50.24 = 200.96 square units

Area of circle = πr² where r is the radius
πr² = 200.96

r² = 200.96/ π = 63.9676 ≅ 64
Therefore r = √64 = 8 units

The circumference of the circle is 2πr = 16π units

arc AB has a vertex angle of 30°.

30°/360° = 1/6

So the length of the arc is 1/6th the circumference of the circle =

(1/6) * 16π  = 2.67π  units or 8.37758 units

3 0
1 year ago
8. If a conditional is true and its hypothesis is true, then its_____ is true.
barxatty [35]
Answer: Conclusion

If a conditional is true and its hypothesis is true, then its conclusion is true.
5 0
2 years ago
Q‒1. [5×4 marks] a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6? (150) b) How many three-
amid [387]

Answer:

a) 294

b) 180

c) 75

d) 174

e) 105

Step-by-step explanation:

I assume that for each problem, the first digit can't be 0.

a) There are 6 digits that can be first, 7 digits that can be second, and 7 digits that can be third.

6×7×7 = 294

b) This time, no digit can be used twice, so there are 6 digits that can be first, 6 digits that can be second, and 5 digits that can be third.

6×6×5 = 180

c) Again, each digit can only be used once, but this time, the last digit must be odd.

If only the last digit is odd, there are 3×3×3 = 27 possible numbers.

If the first and last digits are odd, there are 3×4×2 = 24 possible numbers.

If the second and last digits are odd, there are 3×3×2 = 18 possible numbers.

If all three digits are odd, there are 3×2×1 = 6 possible numbers.

The total is 27 + 24 + 18 + 6 = 75.

d) If the first digit is 3, and the second digit is 3, there are 1×1×6 = 6 possible numbers.

If the first digit is 3, and the second digit is greater than 3, there are 1×3×7 = 21 possible numbers.

If the first digit is greater than 3, there are 3×7×7 = 147 numbers.

The total is 6 + 21 + 147 = 174.

e) If the first digit is 3, and the second digit is greater than 3, then there are 1×3×5 = 15 possible numbers.

If the second digit is greater than 3, there are 3×6×5 = 90 possible numbers.

The total is 15 + 90 = 105.

6 0
3 years ago
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